Is ${974613}$ divisible by $3$ ?
Answer: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {974613}= &&{9}\cdot100000+ \\&&{7}\cdot10000+ \\&&{4}\cdot1000+ \\&&{6}\cdot100+ \\&&{1}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {974613}= &&{9}(99999+1)+ \\&&{7}(9999+1)+ \\&&{4}(999+1)+ \\&&{6}(99+1)+ \\&&{1}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {974613}= &&\gray{9\cdot99999}+ \\&&\gray{7\cdot9999}+ \\&&\gray{4\cdot999}+ \\&&\gray{6\cdot99}+ \\&&\gray{1\cdot9}+ \\&& {9}+{7}+{4}+{6}+{1}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${974613}$ is divisible by $3$ if ${ 9}+{7}+{4}+{6}+{1}+{3}$ is divisible by $3$ Add the digits of ${974613}$ $ {9}+{7}+{4}+{6}+{1}+{3} = {30} $ If ${30}$ is divisible by $3$ , then ${974613}$ must also be divisible by $3$ ${30}$ is divisible by $3$, therefore ${974613}$ must also be divisible by $3$.